Talk:Hill sphere

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Contents 1 Roche lobe 2 Hill sphere of moon 3 Comments, verification 4 Hill sphere and Distance a (part 1) 5 Hill sphere and Distance a (part 2) 6 Hill sphere and Distance a (part 3)

[edit] Roche lobe

Most of the sources I can find seem to indicate that the roche lobe and the hill sphere are really the same thing. (the latter being a spherical approximation of the former?). They both have similar definitions and represent the same concept as far as my understanding goes. However the article seems to say that the two must not be confused. Is anyone able to clarify this? 69.157.226.139 16:55, 18 April 2007 (UTC)

[edit] Hill sphere of moon

Does anyone know if the Moon can support a stable orbit? A quick calculation gives it a Hill sphere of ~60,000 km with respect to Earth, and ~350,000 km with respect to the Sun, however it's not really a 3-body but a 4-body problem in this case. Deuar 21:53, 9 June 2006 (UTC)

[edit] Comments, verification

Article states "It is also called the Roche sphere because the French astronomer Édouard Roche independently described it." My information points to Roche describing it mathematically and the concept merely being borrowed intellectually for the specific purposes intended by Hill, in which event it may indeed be called a Hill sphere or a Roche sphere, but it was not "independently described" by Roche - it was more a "child" of his work. Please let's discuss!

24.215.19.163 11:22, 18 December 2006 (UTC)

[edit] Hill sphere and Distance a (part 1)

In using the Semi-Major Axis in the formula, a literal interpretation would mean distance to the barycenter (shared by the two bodies), not the distance between the actual bodies themselves. Does the author mean the distance to the barycenter (which could be many km outside the larger body)? Or, does the author mean the distance between the two bodies (center to center)? I assume that it is the latter (since the barycenter is the result of the shared orbits). Tesseract501 10:48, 24 February 2007 (UTC)

Well I'm not the author, but I would surmise that the shortest distance between the large and smaller body is most critical. The moment of closest approach is the time when the third orbiting body is most perturbed. Of course since the large body in question is usually the Sun, the difference is insignificant anyway. Especially since we have a ≈ in the formula, indicating that the true practical distance at which bodies orbiting the smaller mass may begin to be lost differs by many percent with respect to the value obtained from the formula. So, the details of whether it's distance to the barycenter or what is lost anyway in general slop and uncertainty. Deuar 14:58, 26 February 2007 (UTC)

[edit] Hill sphere and Distance a (part 2)

Another queston: The formula in this article indicates the value for "a" as the Semi-major Axis. I've seen other sites use the distance of closest approach (Periapsis) in the assumption that the Hill Sphere size changes based on the distance of the two objects. Which is correct? The formula on this page incorporates the eccentricity of the orbit. Does that support using the SMA? What if the eccentricity is high and the smaller, comet-like object is much closer at Periapsis than at Apapsis? If the Hill Sphere does, in fact, shrink at Periapsis, using the SMA in the formula may show a Hill Sphere that is too large to accommodate stable orbits at closest approach. Tesseract501 10:52, 24 February 2007 (UTC)

The formula with the eccentricity is the correct one. Since the loss process involves an acumulation of small kicks at each orbit, it is useless to talk about the Hill sphere changing with location in the orbit. If the body is too far out at any time at all, it will eventually be lost after making enough orbits. This is why the moment when perturbations are strongest (perihelion) is the relevant time at which to calculate the Hill radius. Deuar 14:58, 26 February 2007 (UTC)

[edit] Hill sphere and Distance a (part 3)

One last question, regarding the reference to the moon having to be LESS THAN 7 months within the Hill Sphere. That sentence seems to pop out with no explanation as to what or why (then again, it may be ignorance). Does the sentence tie in to the use of SMA (instead of using the distance of closest approach)? To me the sentence implies a safety period for a stable orbit that is temporarily outside of the Hill Sphere? If necessary, may someone clarify the meaning of the 7-month statement? Tesseract501 10:58, 24 February 2007 (UTC)

This seems to be just someone's way to phrase the fact that all stable orbits around the Earth must have an orbital period of 7 months or less. It does not refer to orbits around any other object. Should be rephrased since it's evidently causing confusion. Deuar 14:58, 26 February 2007 (UTC)